3y^2+16y-32=0

Solution for 3y^2+16y-32=0 equation:

3y^2+16y-32=0

a = 3; b = 16; c = -32;
Δ = b2-4ac
Δ = 162-4·3·(-32)
Δ = 640
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-8\sqrt{10}}{2*3}=\frac{-16-8\sqrt{10}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+8\sqrt{10}}{2*3}=\frac{-16+8\sqrt{10}}{6}
$